How can I solve the integration of 1/(sin^4x+cos^4x) and also 1/sin^6x+cos^6x?
Interesting. Now, there are two questions that are being asked here, so it’s necessary that you bear with me as I answer each one individually.We’re searching for the answer to the following question first;Let us start off by messing around with the integrand a little bit. See, it’s interesting that people often jump to one of the integration methods without trying to screw around with the integrand too much. Where’s the fun in life if you go through such a linear route?Anyways, here you go;Now, it looks way simpler than before. I’ll go ahead and simplify it even further;It should be rather obvious where I am taking this. We can conclude, based on the above manipulations, that;To solve this, let us use the following substitution;If we plug in the above expressions into the integral, we will get;You can clearly see that this integral is very easy. It is elementary;Therefore, we conclude that;That answers the first question. We shall, therefore, swiftly move on to the next question. Let me say, however, that it’s very easy to see me do all of this through this answer. That is, it’s easy to watch me do the math and tell yourself that it looks very reasonable. The only way you’ll find out is by attempting it yourself. I urge you to write out the details in my derivation and to understand it step-by-step. This is for your sake.Let us denote the second integral with;Again, the way we’ll tackle this is by playing around with the integrand just a little bit:Of course, you must realize that dG/dx refers directly to the integrand. I’m just playing around with notation, because I just love messing with people.Anyway, you can see that we can still simplify that massive denominator using the work that we did with the first integral, because a similar expression pops up in the denominator of the above fraction;The junk term is, of course, the following;To simplify this, we notice that there is a common factor and we bring that out;Therefore, our integrand will now be;I hope it is fairly obvious that we can now solve this with the method that we used for the first integral. We can say that;Let’s use the following substitution;Therefore, our integral becomes;Therefore, the answer is;Cheers :)
How do I integrate a hyperbola?
Contrary to appearance, this is a tough integral indeed... ∫ √(x² - 1) dx = substitute x = secθ dx = secθ tanθ dθ yielding: ∫ √(x² - 1) dx = ∫ [√(sec²θ - 1)] secθ tanθ dθ = recall that sec²θ - 1 = tan²θ: ∫ [√(tan²θ)] secθ tanθ dθ = ∫ tanθ secθ tanθ dθ = ∫ tan²θ secθ dθ = replace tan²θ with (sec²θ - 1): ∫ (sec²θ - 1) secθ dθ = expanding it and then breaking it up, ∫ (sec³θ - secθ) dθ = ∫ sec³θ dθ - ∫ secθ dθ thus, summing up: ∫ tan²θ secθ dθ = ∫ sec³θ dθ - ∫ secθ dθ (##) integrate the first integral on the right side by parts (rewriting it as ∫ secθ sec²θ dθ), letting: sec²θ dθ = dv → tanθ = v secθ = u → secθ tanθ dθ = du yielding: ∫ u dv = v u - ∫ v du → (∫ sec³θ dθ =) ∫ secθ sec²θ dθ = tanθ secθ - ∫ tanθ secθ tanθ dθ → ∫ sec³θ dθ = tanθ secθ - ∫ tan²θ secθ dθ plug this into the above (##) expression, obtaining: ∫ tan²θ secθ dθ = ∫ sec³θ dθ - ∫ secθ dθ ∫ tan²θ secθ dθ = tanθ secθ - ∫ tan²θ secθ dθ - ∫ secθ dθ → having the same integral on both sides, collect it on the left side: ∫ tan²θ secθ dθ + ∫ tan²θ secθ dθ = tanθ secθ - ∫ secθ dθ → hence ∫ tan²θ secθ dθ = (1/2)tanθ secθ - (1/2) ∫ secθ dθ → multiply the integrand on the right side by (tanθ + secθ)/(secθ + tanθ) (= 1): ∫ tan²θ secθ dθ = (1/2)tanθ secθ - (1/2) ∫ (tanθ + secθ)secθ dθ /(secθ + tanθ) = expand the top: (1/2)tanθ secθ - (1/2) ∫ (tanθ secθ + sec²θ) dθ /(secθ + tanθ) = note that the top is the derivative of the bottom: (1/2)tanθ secθ - (1/2) ∫ d(secθ + tanθ) /(secθ + tanθ) = yielding: (1/2)tanθ secθ - (1/2) ln |secθ + tanθ| + c now recall that secθ = x hence tanθ = √(sec²θ - 1) = √(x² - 1) thus, substituing back, you end with: ∫ √(x² - 1) dx = (1/2)tanθ secθ - (1/2) ln |secθ + tanθ| + c = (1/2)[√(x² - 1)] x - (1/2) ln |x + √(x² - 1)| + c I hope it helps..
How do I integrate sin2xcos2xdx?
∫sin(2x)*cos(2x) dx u=sin(2x) du/2 = cos(2x) dx 1/2*∫u du =u²/4 + C =sin²(2x)/4 + C
Integrating 1. y= 1/[(cosx)^3 ] ?
Integral ( 1/cos^3(x) dx ) Since cos^3(x) = 1/sec^3(x), the above is the same as Integral ( sec^3(x) dx ) This is not an easy integral to solve, because it uses integration by parts and "going in a circle", as well as knowledge that the integral of sec(x) is ln|sec(x) + tan(x)|. I will demonstrate. Integral ( sec^3(x) dx ) First, I'm going to split this as sec(x) and sec^2(x). Integral ( sec(x) sec^2(x) dx ) Let u = sec(x). dv = sec^2(x) dx. du = sec(x)tan(x) dx. v = tan(x) uv - Integral ( v du ) sec(x)tan(x) - Integral ( tan(x) sec(x) tan(x) dx ) sec(x)tan(x) - Integral ( sec(x) tan^2(x) dx ) Use the identity tan^2(x) = sec^2(x) - 1.l sec(x)tan(x) - Integral ( sec(x) [sec^2(x) - 1] dx ) sec(x)tan(x) - Integral ( [sec^3(x) - sec(x)] dx ) Split into two integrals, sec(x)tan(x) - ( Integral [sec^3(x) dx ] - Integral [sec(x) dx ] ) Distribute the minus. sec(x)tan(x) - Integral [sec^3(x) dx ] + Integral [sec(x) dx ] Evaluate the last integral. sec(x)tan(x) - Integral [sec^3(x) dx ] + ln|sec(x) + tan(x)| Looks like we are back to integrating sec^3(x), right? Wrong, because consider that the above expression is actually an equation. The equation is Integral ( sec^3(x) dx ) = sec(x)tan(x) - Integral [sec^3(x) dx ] + ln|sec(x) + tan(x)| So on the left hand side, we have the integral of sec^3(x), and on the right hand side, the same thing. What we are going to do is move the negated integral of sec^3(x) to the left hand side. This gives us Integral ( sec^3(x) dx ) + Integral ( sec^3(x) dx) = sec(x)tan(x) + ln|sec(x) + tan(x)| And because we have identical terms being added together on the left hand side, that means there are TWO of them. 2 * Integral ( sec^3(x) dx ) = sec(x)tan(x) + ln|sec(x) + tan(x)| And now that we merged together both integrals on the left hand side which puts a constant there as a result, all we have to do is get rid of that constant by multiplying both sides by (1/2). Integral ( sec^3(x) dx ) = (1/2)sec(x)tan(x) + (1/2)ln|sec(x) + tan(x)| We originally wanted to solve for the integral of sec^3(x), and there it is. Oh, and don't forget to add a constant. Integral ( sec^3(x) dx ) = (1/2)sec(x)tan(x) + (1/2)ln|sec(x) + tan(x)| + C
How do I integrate [math]4x^{3}e^{x^4}[/math]?
∫(4x^3 e^x^4) d x4∫(x^3 e^x^4) d xu = x^4du = 4x^3∫(e^u) d ue^u=> e^x^4+C
How do you integrate 1/x^3?
Rewrite it as x^-3 and use the power rule INT x^n dx = (x^(n+1))/(n+1). This works with all fractional and negative powers except n = -1. INT x^-3 dx = (-1/2)x^-2 + c = -1/(2x^2) + c
How can one solve the following by using the method of factorization: [math]\frac{4}{x-3}=\frac{5}{2x+3}[/math]where [math]x[/math] is not equal to [math]0[/math] or [math]-\frac{3}{2}?[/math]
4/x-3=5/2x+3Multiply both sides by (x-3)(2x+3)4(2x+3)=5(x-3)=8x+12=5x-15Subtract 12 from both sides8x+12 -12 =5x-15 -12= 8x=5x-27Subtract 5x from both sides.8x -5x =5x-27 -5x= 3x=-27x=-9
The sum of the first hundred terms of an arithmetic progression with first term a and common difference d is T?
I am assuming that sum of the first 50 odd-numbered terms is (1/2)T - 1000 rather than 1/(2T) - 1000 Next time try to avoid ambiguous fractions in your expressions. Answers doesn't have an easy fractional notation, and we need parentheses to make it clear. Okay, let's look at our arithmetic progression: a, a+d, a+2d, etc. The sum of the first N terms is Na + [0+1+2+ .. + (N-2) + (N-1)]d = Na + N(N-1)d/2 using the formula for triangular numbers to deal with the sum 0+1+2+ .. + (N-2) + (N-1) = N(N-1)/2 For the first 100 terms, this works out to {1} 100a + 4950d = T Now consider the sum of the first 50 odd-numbered terms: a, a+2d, a+4d, a+6d .. But these are just the first 50 terms of an arithmetic progression with first term a and commond difference 2d, and we already have a formula to apply here: 50a + 50 * 49 * 2d /2 = 50a + 2450d = T/2 - 1000 We can double both sides of that last equation and simplify to get another expression for T: {2} 100a + 4900d = T - 2000 Subtracting equation {2} from equation {1} yields 50d = 2000 so d = 2000/50 = 40